Converting kW/h to kWh for Optimizing Fuel Cell Ramp-Up

[Ornella]

Hi everyone,

I am working on a mathematical optimization model for a fuel cell.
Currently I am facing a problem with the ramp-up of the cell.
I have a modulation ramp of 4% of the nominal power (58.3 kW) per minute.
My constraint in the model has to be in kWh (I have to precise that my model is a hourly simulation during one year).
Concerning the conversion of the power in other constraints (as for FC capacity constraint) I simply multiplied the power times one hour and I should get the energy in kWh.
But for the ramp-up constraint I am really struggling with the conversion.
Anyone that has an idea how to obtain it?

Thanks in advance for any help.
Ornella

 

[A.T.]

I simply multiplied the power times one hour and I should get the energy in kWh.

That works for constant power. For variable power you have take its integral over time.

 

[Ornella]

That works for constant power. For variable power you have take its integral over time.

The power is actually constant. I am trying to expand the simulation per every hour of the year. Is it still wrong?

 

[Merlin3189]

I have a modulation ramp of 4% of the nominal power (58.3 kW) per minute.

Maybe:
At the start you are drawing no current. Every minute you can increase your power draw by 4% of 58.3kW (= 2.332kW)
After 15min you are drawing the full rated power of 58.3kW.

 

[Merlin3189]

If you are looking at energy, then, provided the ramp is a straight line (as it usually seems to be),
energy during ramp = average power x time = 1/2 full power * ramp time. = 0.5 x 58.3 x 0.25 kWh

 

[A.T.]

The power is actually constant.

Than what are you measuring in kW/h?

 

[Ornella]

Than what are you measuring in kW/h?

the energy during the ramp up of my fuel cell

 

[Ornella]

If you are looking at energy, then, provided the ramp is a straight line (as it usually seems to be),
energy during ramp = average power x time = 1/2 full power * ramp time. = 0.5 x 58.3 x 0.25 kWh

Thank you very much! That's what I was looking for!

 

[Ornella]

Maybe:
At the start you are drawing no current. Every minute you can increase your power draw by 4% of 58.3kW (= 2.332kW)
After 15min you are drawing the full rated power of 58.3kW.

I actually found out the time for ramp up to be 25 min.

0.04/min * 60 min/h * 58 kW = 139.9 kW/h
58.3 kW / 139.92 = 0.42 h = 25 min
Then I applied the method as you said.
0.5 * 58.3 * 0.42 = 12.5 kWh
Thanks again!

 

[Merlin3189]

Yes you're quite right. My brain must be withering! 25 minutes at 4% per minute is 100%

 

[A.T.]

Then what are you measuring in kW/h?

the energy during the ramp up of my fuel cell

Energy is measured in kWh, not in kW/h.

 

[Ornella]

Energy is measured in kWh, not in kW/h.

I was trying to obtain the energy during the actual time of ramp-up. But I solved my problem.
Thanks anyways.